$\textbf{Determine all function $\mathbf{f:\mathbb{R}\rightarrow\mathbb{R}}$ such that\\\\ $\mathbf{f(x+y)f(x-y)=\left(f(x)+f(y)\right)^2-4x^2f(y)}$,each x,y\in \mathbb{R}$}\\
I have got a quadratic expression.
i am getting 2 solutions :
one is trivial i.e. f(x)=0 for all x element of R
else
we put x=y=0 in the relation
{f(0)}^2=4{f(0)}^2
which implies f(0)=0;
then if we put x=0 we hav
f(y)f(-y)=f(y)^2
since we hav considered f(y) not =0 we hav f(y)=f(-y)
then having put y=-x we hav
f(0)f(2x)=4{f(x)}^2 - 4x^2 f(x)
this implies
f(x)=x^2
1but f(x) = x2 isn't the answer.
check L.H.S = x4 - y4
but R.H.S = (x2 - y2)2
..
21Let P(x,y) be the assertion .
P(0,0) → 3(f(0))2 = 0→ f(0) = 0
P(x,x)→ f(x)(f(x)-x)= 0. This means if f(x)≠0 , f(x)= x2
Let there exist some t (t≠0,obviously) such that f(t)≠0 and hence f(t)= t2 and some m such that f(m) = 0
P(t,m) →f(t+m)f(t-m) = t4,Hence f(t+m) and f(t-m) both can't be zero. therefore f(t-m) = (t-m)2= f(m-t)
P(m,t)→f(m+t)f(m-t) = t4 - 4m2t2 (*) but f(t+m)f(t-m) = f(m+t)f(m-t)=t4
Hence we have from (*) that m2t2= 0 → m= 0(since t≠0)
Therefore if f(x) is nonzero for some x, it must be f(x) = x2 (x≠0) and f(0) = 0, otherwise f(x) is zero for all real x.
So the solutions (which are indeed solutions) are f(x) = x2 and f(x) = 0
21It can be done w/o using the assertion P(m,t) .
P(t,m) (above post) implies (t2-m2)2= t4 which gives
1) m = 0 2)m2= 2t2 but this is not possible...
because let k≠{t , -t ,±t√2} f(k) ≠0 and hence f(k) = k2 , so we must have m2 = 2k2= 2t2 implying t2= k2 but that's contradiction according to our choice of k.
1for x=y=0, you have f(0)=0 ---1)
for x=y you get a quadratic for f(x) in terms of x,
solve to obtain x2 or 0
cheers!
21thats wrong...
P(x,x)→ f(x)(f(x)-x)= 0. This means if f(x)≠0 , f(x)= x2
to say anything more u have to do what i did..
1well this is the INMO 2011 Q6.
Here is the link http://olympiads.hbcse.tifr.res.in/uploads/inmo-sol-2011
